Problem: The equation of a circle $C$ is $x^2+y^2+16x-16y+79 = 0$. What is its center $(h, k)$ and its radius $r$ ?
To find the equation in standard form, complete the square. $(x^2+16x) + (y^2-16y) = -79$ $(x^2+16x+64) + (y^2-16y+64) = -79 + 64 + 64$ $(x+8)^{2} + (y-8)^{2} = 49 = 7^2$ Thus, $(h, k) = (-8, 8)$ and $r = 7$.